In which direction will conventional current flow through the resistor in Figure 20.87? What will be the direction of the magnetic force on the moving bar?

The direction of conventional current is define by using the following formulae.

The current is,

$\mathit{I}\mathbf{=}\frac{\mathbf{\epsilon }}{\mathbf{R}}$

Here, $\epsilon$ Is the voltage and $R$ is the resistor.

The emf voltage is defined by,

$\epsilon =\int {\overrightarrow{f}}_{\text{mag}}\cdot \overrightarrow{dl}$

Here, ${\overrightarrow{f}}_{\text{mag}}$ is the magnetic force and $\overrightarrow{dl}$ is the change in distance.

The magnetic force is,

${\overrightarrow{f}}_{\text{mag}}=\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

Here,$\overrightarrow{v}$ is the velocity of the charge and$\overrightarrow{B}$ is the magnetic field.

From the given figure,

The velocity, $\overrightarrow{v}=v\left(\widehat{x}\right)$

The magnetic field,$\overrightarrow{B}=B\left(+\widehat{z}\right)$

Therefore, the magnetic force will be,

$\begin{array}{c}{\overrightarrow{f}}_{\text{mag}}=\left(\overrightarrow{v}\times \overrightarrow{B}\right)\\ =vB\left(\widehat{x}\times \widehat{z}\right)\\ =vB\left(-\widehat{y}\right)\end{array}$

The magnetic files is directed into the page, and the bar is moving with speed v in the positive x direction.

This situation can be seen in the following figure.

In the above figure, one can see that the magnetic field is directed into the page, and the bar is moving with speed v in the positive x direction. Electrons moves in the downward direction, leaving the top end of the bar positively charged and the bottom end of the bar negatively charged because the magnetic force $\left({\overrightarrow{F}}_{\text{mag}}\right)$ on the mobile electrons in the bar points in the negative y direction. Therefore, the conventional current (positive charge) flows through the resistor from the top end of the bar to the bottom end of the bar.

This means that the conventional current flow through a resistor in the direction of -y axis.

The magnetic force on the conductor is always in the direction so that it opposes the induced emf, which means always in the direction such that it opposes the motion of the conductor.

Hence, the magnetic force on the bar is towards -x axis.